(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Consider the first excited state of the Hydrogen atom. The principle quantum number is given by n = 2 and so it is four-fold degenerate. Consider now a weak perturbation in the form of V = λxy, where x and y are the Cartesian coordinates of the electron with respect to the proton. The explicit forms of the involved spherical harmonics are also given below:

[tex]Y_{00} = \frac{1}{\sqrt{4\pi}}[/tex]

[tex]Y_{10} = \frac{3}{\sqrt{4\pi}}cos(\theta) = \frac{3}{\sqrt{4\pi}}\frac{z}{r}[/tex]

[tex]Y_{11} = -\frac{3}{\sqrt{8\pi}}sin(\theta)e^{i\phi} = -\frac{3}{\sqrt{8\pi}}\frac{x + iy}{r}[/tex]

[tex]Y_{1-1} = \frac{3}{\sqrt{8\pi}}sin(\theta)e^{-i\phi} = \frac{3}{\sqrt{8\pi}}\frac{x - iy}{r}[/tex]

Based on the first-order degenerate perturbation theory, describequalitativelyhow the degenerate levels will shift or split due to the perturbation.

2. The attempt at a solution

In this case of four-fold degeneracy, the first-order corrections of energy E^{1}are given by the eigenvalues of a W-matrix, which has the following elements:

W_{ij}= < ψ_{i}^{0}|V| ψ_{j}^{o}>, i,j = 1,2,3,4

Here for simplicity I denote ψ_{200}, ψ_{21-1}, ψ_{210}, ψ_{211}using ψ_{1}, ψ_{2}, ψ_{3}, ψ_{4}respectively, and ψ_{nlm}stands for the state of quantum number n, l, and m.

Then I use the given spherical harmonics to determine whether certain element of matrix W is zero. For example,

W_{11}= <ψ_{1}^{0}|V|ψ_{1}^{0}> = 0

because the φ part of above equation converges to 0 when integrated from 0 to 2π.

After I went through all the elements of W, I got:

W_{11}= W_{22}= W_{33}= W_{44}= W_{13}= W_{23}= W_{43}= W_{31}= W_{32}= W_{34}= 0

Since I am asked to give a qualitative description, I did not work out the specific values of the other elements, but denote them using:

W_{12}= a_{1}, W_{14}= b_{1}, W_{21}= a_{2}, W_{24}= c_{1}, W_{41}= b_{2}, W_{42}= c_{2}

So finally I have this characteristic equation for W:

λ^{4}- (a_{1}a_{2}- b_{1}b_{2}+ c_{1}c_{2})λ^{2}- (a_{1}b_{2}c_{1}+ a_{2}b_{1}c_{2})λ = 0

Here λ represents the eigenvalues of W and hence is the first-order correction of energy.

Thanks to. I lost the root λ = 0 during my previous calculation.phyzguy

There are totally four degenerate states in the beginning, and only three different energy levels after perturbation. It is said that "typically the perturbation will break or lift the degeneracy", and I do not understand why here I lost one degeneracy. So I was wondering whether my calculations were wrong somewhere, i.e., some of the elements of W may not be zero.

I have been working on this problem for days, calculating around, but still cannot figure it out.

Thanks.

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# Homework Help: Quantum Perturbation Theory

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